Listen carefully. We have five marked points on a 7×7 grid with coordinates (x,y):
- A = (5, 0)
- B = (6, 3)
- C = (1, 3)
- D = (6, 2)
- E = (3, 2)
The path must visit all five points in a single continuous order so that the four consecutive segment lengths equal, in order: 1, √10, 5, √5.
Step 1 — Use the Pythagorean theorem to compute squared distances between each pair. (We compare squared distances to avoid roots.)
- AB: (6−5)^2 + (3−0)^2 = 1 + 9 = 10 → distance = √10
- AC: (1−5)^2 + (3−0)^2 = 16 + 9 = 25 → distance = 5
- AD: (6−5)^2 + (2−0)^2 = 1 + 4 = 5 → distance = √5
- AE: (3−5)^2 + (2−0)^2 = 4 + 4 = 8 → distance = √8
- BC: (1−6)^2 + (3−3)^2 = 25 + 0 = 25 → distance = 5
- BD: (6−6)^2 + (2−3)^2 = 0 + 1 = 1 → distance = 1
- BE: (3−6)^2 + (2−3)^2 = 9 + 1 = 10 → distance = √10
- CD: (6−1)^2 + (2−3)^2 = 25 + 1 = 26 → distance = √26
- CE: (3−1)^2 + (2−3)^2 = 4 + 1 = 5 → distance = √5
- DE: (3−6)^2 + (2−2)^2 = 9 + 0 = 9 → distance = 3
Step 2 — Use elimination. The first required segment length is 1. Look for the pair of points distance 1 apart: only BD has squared distance 1. Therefore the first segment must be between B and D. That means the path starts at D and goes to B, or starts at B and goes to D.
Step 3 — Check which order works with the next required length √10. If the order were B → D as the first segment, then the next segment must start at D and go to some point at distance √10 from D. But D has distances: to A √5, to B 1, to C √26, to E 3 — none are √10. So B → D cannot be followed by √10. Therefore the only possible start is D → B (so the first segment is D to B = 1).
Step 4 — From B (the second point), the next required length is √10. Which points are √10 from B? We have AB = √10 and BE = √10. So the third point must be either A or E.
Try B → A: If the third point is A, the next required length is 5. From A, the only point at distance 5 is C (AC = 5). Then the final required length is √5, and C → E has distance √5 (CE = √5). That gives the path:
D → B → A → C → E
Verify lengths in order:
- D→B: 1 (BD = 1)
- B→A: √10 (AB = √10)
- A→C: 5 (AC = 5)
- C→E: √5 (CE = √5)
Everything matches the required sequence 1, √10, 5, √5. The other possibility (B → E as the third point) fails because E has no neighbor at distance 5. So the unique valid path (with the required order starting with length 1) is:
D (6,2) → B (6,3) → A (5,0) → C (1,3) → E (3,2)
Teacher comments (Amy Chua cadence):
You did the right thing by computing distances with the Pythagorean theorem — good. But do not be sloppy: write every squared-distance before taking a square root so you don’t make silly arithmetic mistakes. I expect neat, logical elimination steps. You found the unique path because you used the single distance-1 pair and forced the order — excellent. If you got it wrong the first time, redo the computations until they are perfect.
Rubric (strict, of two levels):
- Exemplar (full credit): Lists all five points, computes all necessary squared distances using the Pythagorean theorem, uses logical elimination (shows why BD must be the 1-length segment), finds the exact path D→B→A→C→E, verifies each consecutive length matches 1, √10, 5, √5, and explains uniqueness. Clear, neat, no arithmetic errors.
- Proficient (partial/full credit): Correct final path D→B→A→C→E with correct verification of the four lengths, but a shorter justification (for example, only computes the needed distances rather than all pairwise distances). Minor omissions in explanation but no incorrect arithmetic.
If you want, draw the 7×7 grid and plot the points A,B,C,D,E, then trace the path D→B→A→C→E to visualize it. Now go practice more: precision and clear work are not optional.